Monotonicity of $\ln n\over  \sqrt{n}$ Is it possible to prove the monotonicity of $(a_n)_n =\frac{\ln n}{\sqrt n}$ without using derivatives?
 A: The sequence $\left\{\dfrac{\ln{n}}{n}\right\}_{n \in \mathbb{N}, \;n \geqslant{3}}$ is monotonically decreasing: $$\dfrac{\ln{n}}{n}-\dfrac{\ln{(n+1)}}{n+1}=\dfrac{(n+1)\ln{n}-n\ln{(n+1)}}{n(n+1)}= \dfrac{\ln{n^{n+1}}-\ln{(n+1)^n}}{n(n+1)}=\dfrac{\ln\dfrac{n^{n+1}}{(n+1)^n}}{n(n+1)}=\dfrac{-\ln\dfrac{(n+1)^n }{n^{n+1}}}{n(n+1)}=-\dfrac{\ln\left[\left(1+\dfrac{1 }{n}\right)^n \cdot\dfrac{1}{n}\right]}{n(n+1)}>0$$ 
for $n>3.$
The last inequality follows from the well-known estimate $${2} \leqslant \left(1+\dfrac{1 }{n}\right)^n \leqslant {3} \quad  \forall n \in \mathbb{N},$$ so that for $n>3$ $$\left(1+\dfrac{1 }{n}\right)^n \cdot\dfrac{1}{n}<1.$$  
A: For $a_n=\dfrac{\ln{n}}{\sqrt{n}}$ we denote $f_n=\dfrac{1}{a_n^2}=\dfrac{n}{\ln^2{n}}.$  In order to prove that $\left\lbrace a_n \right\rbrace $ is decreasing, we consider 
\begin{gather*}
f_{n+1}-f_n=\dfrac{n+1}{\ln^2(n+1)}- \dfrac{n}{\ln^2{n}}=
\dfrac{(n+1)\ln^2{n}-n\ln^2(n+1)}{\ln^2{n} \ \ln^2(n+1)}=\dfrac{A}{B}
\end{gather*}
and show that sequence $\left\lbrace f_n  \right\rbrace$ increases: $f_{n+1}-f_n >0 \quad \left(\forall n \geqslant n_0 \right).$
Numerator $A$ may be rewritten as
\begin{gather*}
A=(n+1)\ln^2{n}-n\ln^2(n+1)=(n+1)\ln^2{n}-n\ln^2\left(n\left(1+\dfrac{1}{n}\right)\right)=\\
(n+1)\ln^2{n}-n\left(\ln{n}+\ln\left(1+\dfrac{1}{n}\right)\right)^2=\\
n \ln^2{n}+\ln^2{n}-n \ln^2{n}-2n \ln{n}\ln\left(1+\dfrac{1}{n}\right)-n \ln^2\left(1+\dfrac{1}{n}\right)=\\
\ln^2{n}-n \ln{n}\ln\left(1+\dfrac{1}{n}\right)-n \ln{n}\ln\left(1+\dfrac{1}{n}\right)-n \ln^2\left(1+\dfrac{1}{n}\right)=\\
\ln{n}\left(\ln{n}-n\ln\left(1+\dfrac{1}{n}\right)\right)-n\ln\left(1+\dfrac{1}{n}\right)\left( \ln{n}+\ln\left(1+\dfrac{1}{n}\right)\right)=\\
\ln{n}\left(\ln{n}-\ln\left(1+\dfrac{1}{n}\right)^n\right)-\ln\left(1+\dfrac{1}{n}\right)^n\left( \ln{n}+\ln\left(1+\dfrac{1}{n}\right)\right).
\end{gather*}
Next, applying the estimate 
$${2} \leqslant \left(1+\dfrac{1 }{n}\right)^n \leqslant {3} \quad  \left(\forall n \in \mathbb{N}\right),$$
we obtain
\begin{gather}
\ln{n}\left(\ln{n}-\ln\left(1+\dfrac{1}{n}\right)^n\right) \geqslant 
\ln{n}\left(\ln{n}-\ln{3} \right);\\
-\ln\left(1+\dfrac{1}{n}\right)^n\left( \ln{n}+\ln\left(1+\dfrac{1}{n}\right)\right) \geqslant -\ln{3}(\ln{n}+\ln{2}).
\end{gather}
Thus, by adding these inequalities,
$$
A \geqslant {\ln{n}\left(\ln{n}-\ln{3} \right) -\ln{3}(\ln{n}+\ln{2})}=
\ln^2{n}-2\ln{3} \ \ln{n}-\ln{2} \ \ln{3}.$$
Quadratic polynomial
$
x^2-2\ln{3}\cdot x-\ln{2}\ln{3}>0 $ for $x\in\left(-\infty,\, x_1\right)\cup{\left(x_2, \,+\infty\right)},
$ where
\begin{gather*}x_1=\ln{3}- \sqrt{\ln^2{3}+\ln{2}\ln{3}}, \\
x_2=\ln{3}+ \sqrt{\ln^2{3}+\ln{2}\ln{3}}.
\end{gather*}
Thus, $A>0$ for $\ln{n}>\ln{3}+ \sqrt{\ln^2{3}+\ln{2}\ln{3}}\approx 2.501626534,$ respectively $n\geqslant n_0= \left \lfloor{e^{2.501626534}}\right \rfloor+1=\left\lfloor {12.20232533}\right\rfloor+1=13.$
This value is less precise than exact value $e^2,$ which is  point of minimum for $f(x)=\dfrac{x}{\ln^2{x}}$.
